Exam 1: Midterm

Solutions and Commentary

Part of the homework for 22C:60, Fall 2007
by Douglas W. Jones
THE UNIVERSITY OF IOWA Department of Computer Science

Grade Distribution

Exam

Mean   =  6.49             X
                     X     X X
                     X     X X X   X  
                     X     X X X X X X
                     X   X X X X X X X
_______________X___X_X_X_X_X_X_X_X_X_X_______
  0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10

Machine Problem 1

Mean   =  4.95

Homework 1 to 5

mean   = 10.67          X
                        X X X  
                        X X X  
                      X X X X  
                      X X X X X
      X           X   X X X X X
______X_X_X_X___X_X_X_X_X_X_X_X__
 0 . 2 . 4 . 6 . 8 . 10. 12. 14.

Overall Midterm Score

mean   = 21.53                                      X  
median = 23.7                                     X X  
                                                  X X  
                                              X X X X  
                                              X X X X  
                    X                   X   X X X X X X
______________X_____X_____X_____X_X_X___X_X_X_X_X_X_X_X_____
 0 . 2 . 4 . 6 . 8 . 10. 12. 14. 16. 18. 20. 22. 24. 26. 28
Rough Grades                F     D     C     B     A

Exam and Solutions

Fill in the following table; values in the blanks in each column derive from the given values in that column, all values in each row are derived from the given values using the same rules. (4 points)

Bit Pattern x         101111      010101      011001     110011     001101

Unsigned decimal     ___47___    ___21___    ___25___      51      ___13___

1's complement of x  _010000_    _101010_     100110    _001100_   _110010_ 

2's complement of x  _010001_     101011     _100111_   _001101_   _110011_

x as a 1's comp num    -16       ___21___    ___25___   __-12___   ___13___

x as a 2's comp num  __-17___    ___21___    ___25___   __-13___      13

Well over half the class did perfectly here. 10 students made 3 or fewer errors, while 4 made 5 or more errors. The most common errors had to do with problems with negative numbers, with 1's complement being more difficult than 2's complement.

Translate the following brief useless SMAL code to a sequence of bytes loaded in memory. The spaces to the right are labeled with a word and byte in word addresses; fill in the hexadecimal value that goes in that byte; leave uninitialized bytes blank. The result of assembling the first lines are given. (2 points)

    . = 0                       ||  Address    3    2    1    0
        W     #01234567         ||          |----|----|----|----|
    ; --solved to here--        ||  000000: | 01 | 23 | 45 | 67 |
        B     A,"A",#A          ||          |----|----|----|----|
        ALIGN 4                 ||  000004: |    | 0A | 41 | 08 |
    A:  B     A,B,C,D           ||          |----|----|----|----|
    B:  H     #ABCD             ||  000008: | 10 | 0E | 0C | 08 |
        ALIGN 2                 ||          |----|----|----|----|
    C:  H     #0123             ||  00000C: | 01 | 23 | AB | CD |
        ALIGN 4                 ||          |----|----|----|----|
    D:  W     #01234567         ||  000010: | 01 | 23 | 45 | 67 |
        END                     ||          |----|----|----|----|
                                ||  000014: |    |    |    |    |
                                            |----|----|----|----|

Only 3 did perfectly here, while 1 earned no credit. 13 earned more than 1.5 points, 10 earned more than 1 point, but less than 1.5. The most common error had to do with byte order. Unfortunately, many students confused "A" (the ASCII character), A (the label) and #A (the hexidecimal value). Another common problem had to do with alignment.

a) Hand assemble the Hawk program fragment given below, assuming the first instruction goes in location 1000₁₆, giving the result as a sequence of halfwords in base 16 and using question marks for unknown values. The first instruction is done for you. (1.5 points)
```
                                    ||         001000:   ??E8
            LIL     R8,ARRAY        ||
        ; --solved to here--        ||         001002:   ????
            LIS     R4,0            ||
        LOOP:                       ||         001004: __00D4__
            LOADSCC R3,R8           ||
            BZS     ENDLOOP         ||         001006: __C8F3__
            ADDSI   R8,4            ||
            ADD     R4,R4,R3        ||         001008: __0302__
            BR      LOOP            ||
        ENDLOOP:                    ||         00100A: __C418__
                                    ||
                                    ||         00100C: __4334__
                                    ||
                                    ||         00100E: __FB00__
```
One student left this blank. Half the class had problems with byte order. The branch instructions posed special problems for many.

b) What does this do? (The answer is very short.) (0.5 points)
```
__Sum_the_words_of_an_array_up_to_the_first_zero_______________________________
```
Half the class tried to give answers that were short and to the point. Three of these were correct, while others omitted or made mistakes with key aspects such as summation, loop termination, or the fact that the array elements were words. 6 left this blank. Most of the remaining answers were not in the form of explanations, but rather, restatements of the computation in a slightly higher level language. In many cases, these attempts were longer than the original program.

The following SMAL Hawk program is related to MP2 and it contains several errors. The comments are correct. The program worked before it was carefully broken. All of the errors were common errors people made on MP2. Find them and suggest corrections. (2 points)

            TITLE   "Similar to MP2, find and print max key in ARRAY"
            USE     "hawk.macs"
            USE     "monitor.h"
            EXT     UNUSED
            EXT     ARRAY
S START

    ; record structure for an object
    STRING  =       0
    KEY     =       4

    ; main program
    START:
            LIL     R2,UNUSED
            LIL     R1,DSPINI
            JSRS    R1,R1           ; end of prologue to program

LIL         LOAD    R8,ARRAY        ; R8 points to the array A
            LIS     R9,0            ; R9 = i = 0 is the loop index
            LIS     R10,0           ; R10 = max = is the largest element
    LOOP:                           ; for (;;) {
            MOVE    R1,R9
            ADDSL   R1,R8,3         ;     R1 is address of A[i]
LOAD        ADDI    R4,R1,KEY       ;     R4 = A[i].key
            LOADCC  R3,R1,STRING    ;     R3 = A[i].string
A[i].string BZS     ENDLOOP         ;     if (A[i].key == NULL) break;
            CMP     R4,R10          ;     if (A[i].key > max)
            BLE     ENDIF           ;       {
MOVE        LOADS   R10,R4          ;         max = A[i].key
    ENDIF:                          ;       }
            ADDSI   R9,1            ;     i = i + 1
            BR      LOOP
    ENDLOOP:                        ; }
MOVE        LIL     R3,R10          ; parameter 1 = max
            LIS     R4,1            ; parameter 2 = 1
            LIL     R1,DSPDEC
            JSRS    R1,R1           ; call putdec( max, 1 )
            LIL     R1,EXIT
            JSRS    R1,R1
            END

The notes in the right margin above give the corrections, a missing specification of the starting address and 4 incorrect instruction names.

13 earned no credit. Only one student noticed the missing starting address. Half the class correctly noticed the LOAD instruction that should have been LIL. About 8 students noticed each of the other mistakes.

Because of the mistake in one comment, grading on that part of the code was lenient, but many students found large numbers of errors that were not there, in some cases, leading to attempts to completely redesign the control structures or register usage of the program.